The structure of the triiodide ion places a negative formal charge on the central iodine atom. The molecular geometry is also linear (at least according to VSEPR), and its electronic geometry is trigonal bipyramidal. Given the high amount of symmetry, shouldn't triiodide be non-polar? Sure, it might have a charged central atom, but so does carbon dioxide.

$\begingroup$$D_{\infty\mathrm{h}}$ molecules like the triiodide ion cannot be polar. Polarity in the context of chemistry would mean having a dipole moment so imo the monopole is irrelevant$\endgroup$
– orthocresol♦Dec 9 '15 at 22:49

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$\begingroup$@orthocresol You should know better than to describe polarity only with dipole of molecule.$\endgroup$
– MithoronDec 9 '15 at 22:55

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$\begingroup$@Mithoron this is going into semantics already - if you consider triiodide polar because each individual I-I bond is polar, then you must also consider CO2, BF3 and CCl4 to be polar. In all cases there is a partial positive charge on the central atom.$\endgroup$
– orthocresol♦Dec 9 '15 at 22:59

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$\begingroup$Yes, polarity isn't well defined and I'm not even sure if it makes sense to ask about polarity of anion.$\endgroup$
– MithoronDec 9 '15 at 23:08

3 Answers
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The structure of the triiodide ion places a negative formal charge on the central iodine atom.

No it doesn’t. The two resonance structures that describe the four-electron three-centre bond put the negative formal charge on the outer iodines ($\ce{1/2-}$ each).

That said, polarity is usually defined as having a non-zero dipole moment. The dipole moment’s vector must display the same symmetry as the entire molecule. Since the molecule is linear and both $\ce{I\bond{...}I}$ distances are equal, its point group is $D_{\infty \mathrm{h}}$ which includes $i$ which means the overall dipole moment must be zero.

The charge it carries does not matter. Any single-atom ion also has zero dipole moment and would thus be called non-polar. ‘Non-polar’ does not mean ‘free of electrostatic interaction’.

$\begingroup$Can you include the resonance structures you are talking about? I have a hard time seeing the formal charge being on the terminal iodine.$\endgroup$
– Martin - マーチン♦Dec 11 '15 at 4:13

$\begingroup$@Martin-マーチン I can’t draw it atm, but it is: $\ce{I-I}~\ce{I-}$$\endgroup$
– JanDec 18 '15 at 22:54

$\begingroup$I see. I did not think about this. So the resonance you are talking about would be $$\ce{I-I\bond{<-}I- <=> {}^{-}I\bond{->}I-I}?$$ But I think we shall not forget about other resonance structures, too. $$\ce{{}^{-}I\bond{->}\overset{+}{I}\bond{<-}I- <=> I-{I- }-I}$$$\endgroup$
– Martin - マーチン♦Dec 22 '15 at 4:25

$\begingroup$@Martin-マーチン And I disagree with the last one because it creates a dectet on the central iodione.$\endgroup$
– JanDec 23 '15 at 0:51

$\begingroup$You realize that all those resonance contributors violate the octet rule, at least when you want to describe their bonding as covalent.$\endgroup$
– Martin - マーチン♦Dec 24 '15 at 6:15

$\begingroup$According to the IUPAC: A molecule is an electrically neutral entity, which would exclude the $\ce{I3-}$, even in Atkin's definition. I made the same mistake though. For the rest of your statements I am happy to give you a thumbs up.$\endgroup$
– Martin - マーチン♦Dec 22 '15 at 4:08

$\begingroup$@Martin-マーチン I wouldn't say Atkins is limiting the word "polar" to molecules. Instead, he is only defining the term "polar molecular" rather than "polar". The journal Inorganic Chemistry seems to think it's ok to refer to polar ions, "The Polar [WO2F4]2- Anion in the Solid State", chemgroups.northwestern.edu/poeppelmeier/pubs/HT/… .$\endgroup$
– DavePhDDec 31 '15 at 12:10

I would argue it is a polar molecular ion.‡ In the comments to this question there has been already pointed out that the term "polarity" is quite ill-defined.
Many chemists understand that if a molecule has a vanishing dipole moment, the molecule itself is non-polar; at least this is true since molecules are neutral by definition. If you actually can extend that definition to charged species is open for debate. I personally dislike this definition - molecule or not.

In this framework carbon dioxide is non-polar, but it reacts with polar substrates, like Grignard agents. The reason for this is its quadrupole moment.
A much more general definition of the term polarity hence should include any molecule that has a multipole of any kind. We can then even include ions.

A polar molecular entity is a molecule that possesses a multipole.

Since the triiodide anion has a non-zero quadrupole moment, I would consider it polar. This point has been and will always be open for debate.

A quick DF-BP86/def2-SVP calculation determines the natural atomic charges as $q(\ce{I_{\mu}})=-0.11$ (the atom in the middle) and $q(\ce{I_{term}})=-0.44$ (the terminal two; deviations from $1.00$ due to rounding). The same calculation finds:

$\begingroup$♦ - just to make sure, your calculations show that the central iodide ion has a -0.11 charge, correct?$\endgroup$
– DissenterDec 18 '15 at 11:35

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$\begingroup$@Martin if you define "polar" as having any multipole, you have to consider H-H, N-N, and every molecule as polar. "It is true, that for H2, N2 etc., wave mechanics, too, gives at least quadrupoles." Fritz London pubs.rsc.org/en/content/articlepdf/1937/tf/tf937330008b$\endgroup$
– DavePhDDec 18 '15 at 13:04

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$\begingroup$@DavePhD I would not dismiss that thought. But what does that add to the specification of the molecule any way.$\endgroup$
– Martin - マーチン♦Dec 18 '15 at 13:09

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$\begingroup$@Martin-マーチン Well, if you define polar as having a permanent electric dipole moment, it specifies some subset of molecules. On another note, is triiodide really linear? "Isolated Triiodide and Pentaiodide Anions..." is saying bent 174.8 degrees onlinelibrary.wiley.com/doi/10.1002/zaac.200800115/abstract$\endgroup$
– DavePhDDec 18 '15 at 13:20

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$\begingroup$@DavePhD All I am trying to say is that looking at the dipole moment for clarification is not enough. And I am saying that that classification does not add anything to the discussion anyway. It's too simple. Also, we might not want to have this discussion in the comment section. Maybe chat, but I am on the road tonight....$\endgroup$
– Martin - マーチン♦Dec 18 '15 at 13:30